This article claims $Ax \times Ay = \det(A)(x\times y)$, but using the vectors $(3,2,1)$ and $(5,4,6)$ and the orthogonal transformation with determinant $1$:
$$\left(\begin{matrix} \frac{1}{2} & \frac{-\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{matrix}\right)$$ does not hold.
$$\left(\begin{matrix} \frac{1}{2} & \frac{-\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{matrix}\right)\left(\begin{matrix} 3 \\ 2 \\ 1 \end{matrix}\right)=\left(\begin{matrix} \frac{-2*\sqrt{3}+3}{2} \\ \frac{3*\sqrt{3}+2}{2} \\ 1 \end{matrix}\right)$$
$$\left(\begin{matrix} \frac{1}{2} & \frac{-\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{matrix}\right)\left(\begin{matrix} 5 \\ 4 \\ 6 \end{matrix}\right)=\left(\begin{matrix} \frac{-4*\sqrt{3}+5}{2} \\ \frac{5*\sqrt{3}+4}{2} \\ 6 \end{matrix}\right)$$
and $$\left(\begin{matrix} \frac{-2*\sqrt{3}+3}{2} \\ \frac{3*\sqrt{3}+2}{2} \\ 1 \end{matrix}\right)\times \left(\begin{matrix} \frac{-4*\sqrt{3}+5}{2} \\ \frac{5*\sqrt{3}+4}{2} \\ 6 \end{matrix}\right)=\left(\begin{matrix} \frac{13*\sqrt{3}+8}{2} \\ \frac{8*\sqrt{3}-13}{2} \\ 2 \end{matrix}\right)$$
but $(3,2,1)\times (5,4,6)=(8,-13,2)$.
So, what does it mean for the cross product to be invariant under orthogonal transformation?
If $v=(3,2,1)$ and $w=(5,4,6)$, then, indeed $v\times w=(8,-13,2)$. And$$A.(v\times w)=\left(4+\frac{13\sqrt{3}}2,4\sqrt{3}-\frac{13}2,2\right).$$It turns out that\begin{align}(Av)\times(Aw)&=\left(\frac{3}{2}-\sqrt{3},1+\frac{3 \sqrt{3}}{2},1\right)\times\left(\frac{5}{2}-2 \sqrt{3},2+\frac{5 \sqrt{3}}{2},6\right)\\&=\left(4+\frac{13\sqrt{3}}2,4\sqrt{3}-\frac{13}2,2\right).\end{align}As you can see, $(Av)\times(Aw)=\det(A).A(v\times w)$ in this case. Actually, it's true for every orthogonal matrix $A$.